Transcript Chapter 8 powerpoint presentation

Ch 8: Stars & the H-R Diagram
 Nick Devereux 2006
Revised 9/12/2012
Spectroscopy
Stellar Spectra
Emission & Absorption Lines
The Sun’s Spectrum
The Spectrum of Hydrogen
Electronic Transitions in the
Hydrogen Atom
Energy Level Diagram
Two important facts about
Hydrogen
1. The ionization potential = 13.6 eV
2. The wavelength of the Ha emission line is 6563Å
where 1Å = 10-10 m
You can figure out everything about Hydrogen from these two
facts and knowing that the energy difference between two
electronic states, DE, is proportional to
DE
a
1/n2
Where n is the principal quantum number
For Example
The Ha absorption line results from the electron jumping from
the n=2 to n=3 level. We can use this fact (#2) to calculate
the constant of proportionality, R
DE = hc/l= constant [ 1/ n12 – 1/n22 ]
So that,
1/l = R [ 1/ n12 – 1/n22 ]
Substitute n1= 2 and n2 = 3 and l = 6563Å
to yield R = 1.097 x 10-3
which is known as the Rydberg Constant
Now you can calculate the
wavelength for all other
electronic transitions
Since,
1/l = 1.097 x 10-3 [ 1/ n12 – 1/n22 ]
But, remember, that this equation yields l in Å
Question: what is the wavelength of the Lyman a transition ?
Question: what is the wavelength that corresponds to the ionization
potential of hydrogen ?
Ionic Spectra
The wavelengths of H like ions, such as He II, Li III, O VIII,
can be estimated using the Bohr model of the H atom;
E = -  U/2
E  Z2 R[ 1/ n12 – 1/n22 ]
So, the energy levels scale by Z2 compared to H.
e.g. He has Z=2 , Z2 = 4, so the Lyman a transition in He I
may be calculated using;
1/l = Z2 1.097 x 10-3 [ 1/ n12 – 1/n22 ]
yielding l = 304 Å which occurs in the far UV part of the spectrum.
Back to the Sun
The Balmer Discontinuity
The Balmer discontinuity is the break in the spectrum at
3646 Å due to the ionization of Hydrogen from the n=2
state;
1/l = 1.097 x 10-3 [ 1/ n12 – 1/n22 ]
Substitute n1 = 2, n2 = ∞, to yield l = 3646 Å
Summary
Stars are made mostly of Hydrogen. So, the stellar continuum
exhibits Hydrogen absorption lines which modifies it from a
perfect Planck function.
The absorption line strengths depend on two things;
How many of the H atoms are in each of the excitation levels,
n=1, n=2, etc, described by the Boltzmann equation
and secondly,
how many of the H atoms are completely ionized, for if the
H atoms are ionized they can not produce any absorption lines.
This is described by the Saha equation.
The Spectral Classification of
Stars
The relative strength of the absorption lines depends on the
temperature of the star. If the star is too hot, all the H is ionized
and there are no absorption lines.
If the star is too cool, all the H atoms will be in the ground state, for
which there are no transitions in the optical part of the spectrum.
A star of “medium” temperature will have lots of H absorption lines.
The range of H absorption lines strengths described above defines
another way to measure stellar temperature.
The Strengths of Absorption
Lines
Boltzmann & Saha Equations
The strength of the absorption lines can be calculated fairly easily
for H and with some difficulty for other species.
The calculation is a two step process.
Step 1 utilizes the Boltzmann equation to determine the relative
number of atoms in the various energy levels with principle
quantum level n =1, n=2, etc.
Step 2 utilizes the Saha equation to determine the relative number
of atoms that are ionized.
Boltmann Equation
N2 = g2 e –h/kT
N1 g1
Where:
N1 = number of atoms in n = 1 level
N2 = number of atoms in n = 2 level
g1 = 2 (statistical weight = degeneracy = 2n2)
g2 = 8
T = temperature
 = frequency
Application of the Boltzmann Equation
For the purposes of this class, we are interested in the Balmer
series of H lines that originate from the n=2 level and occur in
the optical part of the spectrum.
The graph in the textbook, Fig. 8.7, shows the relative number
of atoms in the n=2 level, as a function of temperature.
The graph shows that very high temperatures are required for
significant population of the n=2 level. Why then does the
strength of the Balmer lines get weaker as the temperature
increases? ie. Why are the Balmer lines so weak in O stars?
The answer is, the H atoms are mostly ionized………
Saha Equation
NII = 2 A (kT)3/2 ZII
___
_________________
NI
e –h/kT
ne ZI
Where=
NI = number of neutral atoms
NII = number of ionized atoms
ZI = partition function for neutral atom (= 2 for H in
gd. State)
ZII = partition function for ionized atom (=1 for H)
ne= electron number density ( e/m3)
A is a constant = 2 me/h3
Application of the Saha Equation
The graph in the textbook, Fig. 8.8, shows the fraction of atoms
that are ionized as a function of temperature.
The graph shows that H becomes ionized at relatively low
temperatures corresponding to ~ 104 K.
The combination of these two equations; Boltzmann and Saha is
illustrated in Fig. 8.9 which shows that the population of the n=2
level is sharply peaked at ~ 104 K. So, the Balmer absorption lines
are strong only in the atmospheres of stars with temperatures ~ 104
K. (see previous figure on slide 17).
The Harvard Spectral sequence
O, B, A, F, G, K, M
The spectral sequence is a much more
precise way to measure the
temperature of stars than the B-V
color index, discussed previously.
The reason is because the B-V color
index is directly affected by the stellar
absorption lines which leads to an
“incorrect” temperature, since stars
are not perfect blackbodies after all.
The Hertzsprung-Russell Diagram
This diagram requires a
measurement of the distance
to each star ( to get Mv) and
a spectrum for each star
( to get the Spectral type ).
Two More Methods to Measure
Distances to Stars
Spectroscopic Parallaxes
The name of this method is a bit deceiving as it incorrectly
implies some measure of stellar parallax. Actually, what happens
is we obtain a spectrum for a star of unknown distance. We use the
spectrum to determine the spectral type, which locates it on the
x-axis of the H-R diagram. Now draw a line up to the main sequence,
and continue it horizontally to determine it’s absolute magnitude, Mv.
The absolute magnitude combined with the apparent magnitude, mv,
allows the distance to be determined using
mv - Mv = 5 log d - 5.
The Second Method is called
Main Sequence Fitting
In this method, we plot a B-V color vs. apparent magnitude
diagram for a cluster of stars of unknown distance.
This graph is then shifted vertically over a copy of a
calibrated H-R diagram until the two main sequences
overlap.
The difference between the cluster apparent magnitude
and the calibrated H-R diagram absolute magnitude is
the distance modulus mv - Mv which yields the distance to
the cluster using mv - Mv = 5 log d - 5.
The M-K Luminosity
Classification scheme